For a fixed $\epsilon > 0$, I want to show that almost surely (i.e., with probability $1$), a standard Brownian motion $W_t$ would change sign over $[0,\epsilon]$.
I thought about defining a random variable $U_t = \mbox{sign} (W_t)$ which would satisfy $(1+U_t)/2$ follows symmetric Bernoulli distribution But I can't conclude with this. Am I on a good track?
On
Another way to solve this problem is with the Blumenthal zero-one law. Let $\mathcal{F}_t = \sigma(B_s : 0 \le s \le t)$, and $\mathcal{F}_{0+} = \bigcap_{t > 0} \mathcal{F}_t$. The Blumenthal zero-one law asserts that $\mathcal{F}_{0+}$ is almost trivial; for each $A \in \mathcal{F}_{0+}$ we have $\mathbb{P}(A) = 0$ or $1$. Consider the event $A_+$ that $B_{1/n} > 0$ for infinitely many $n$. Show that $\mathbb{P}(A_+) \ge 1/2$, and that $A_+ \in \mathcal{F}_{0+}$. Conclude that $\mathbb{P}(A_+) = 1$. Show the same for $A_-$, the event that $B_{1/n} < 0$ for infinitely many $n$. On the event $A_+ \cap A_-$, $B_t$ changes sign in every interval $[0,\epsilon]$.
The reflection principle says that for any $\epsilon>0$ and $x>0$, we have $$\mathbb{P}_0(\sup_{0\leq t\leq \epsilon} W_t \geq x)=2\mathbb{P}_0(W_\epsilon\geq x)= 2\mathbb{P}\left(Z\geq x/\sqrt{\epsilon}\right). $$ Letting $x\downarrow 0$ we conclude that $$\mathbb{P}_0\left(\sup_{0\leq t\leq \epsilon } W_t >0 \right)=2\mathbb{P}(Z> 0)=1.$$ In any time interval, no matter how short, the Brownian motion $(W_t)$ almost surely takes a strictly positive value. Applying this to the Brownian motion $(-W_t)$, we see that $(W_t)$ also almost surely takes a strictly negative value. So the process $(W_t)$ changes sign with probability one.